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A233307
a(n) = |{0 < k < n: p(k)^2 + q(n-k)^2 is prime}|, where p(.) is the partition function (A000041) and q(.) is the strict partition function (A000009).
15
0, 1, 2, 2, 1, 1, 4, 2, 3, 2, 2, 4, 4, 3, 2, 2, 5, 3, 1, 5, 3, 5, 6, 3, 3, 2, 2, 1, 1, 2, 2, 5, 3, 4, 3, 5, 3, 1, 6, 4, 7, 10, 3, 5, 4, 2, 4, 5, 3, 4, 2, 3, 7, 9, 5, 6, 8, 2, 5, 3, 3, 5, 4, 3, 5, 4, 6, 7, 6, 3, 2, 9, 8, 6, 1, 6, 7, 7, 6, 2, 5, 8, 4, 6, 2, 6, 4, 8, 7, 3, 5, 3, 3, 5, 4, 5, 8, 5, 6, 2
OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1. Also, for any integer n > 4, p(k)*q(n-k) - 1 is prime for some 0 < k < n/2.
(ii) If n > 9, then prime(k)*p(n-k) + 1 is prime for some 0 < k < n. If n > 2, then prime(k)*q(n-k) - 1 is prime for some 0 < k < n, and also prime(k)*q(n-k) + 1 is prime for some 0 < k < n.
(iii) If n > 11, then prime(k) + p(n-k) is prime for some 0 < k < n. If n > 4, then prime(k) + q(n-k) is prime for some 0 < k < n, and also prime(k)^2 + q(n-k)^2 is prime for some 0 < k < n.
LINKS
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.
EXAMPLE
a(5) = 1 since 5 = 1 + 4 with p(1)^2 + q(4)^2 = 1^2 + 2^2 = 5 prime.
a(6) = 1 since 6 = 3 + 3 with p(3)^2 + q(3)^2 = 3^2 + 2^2 = 13 prime.
a(19) = 1 since 19 = 3 + 16 with p(3)^2 + q(16)^2 = 3^2 + 32^2 = 1033 prime.
a(28) = 1 since 28 = 3 + 25 with p(3)^2 + q(25)^2 = 3^2 + 142^2 = 20173 prime.
a(29) = 1 since 29 = 6 + 23 with p(6)^2 + q(23)^2 = 11^2 + 104^2 = 10937 prime.
a(38) = 1 since 38 = 1 + 37 with p(1)^2 + q(37)^2 = 1^2 + 760^2 = 577601 prime.
a(75) = 1 since 75 = 13 + 62 with p(13)^2 + q(62)^2 = 101^2 + 13394^2 = 179409437 prime.
a(160) = 1 since 160 = 48 + 112 with p(48)^2 + q(112)^2 = 147273^2 + 1177438^2 = 1408049580373 prime.
a(210) = 1 since 210 = 71 + 139 with p(71)^2 + q(139)^2 = 4697205^2 + 8953856^2 = 102235272080761 prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[PartitionsP[k]^2+PartitionsQ[n-k]^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 07 2013
STATUS
approved